## Dividing a Segment into N parts:

Similar Right Triangles

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Copyright © 1996-2018 Alexander Bogomolny### Dividing a Segment into N parts:

Similar Right Triangles

Following is a simple algorithm for finding the N^{th} part of a given segment,

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Let AB be a given segment and C a point on the line AB satisfying

Indeed, triangles AEC and APE are right and share an angle at A. They are therefore similar implying a proportion

(1) | AP / AE = AE / AC |

However, AC = N·AB = N·AE. Thus (1) gives

(2) | AP = AE² / AC = AE / N = AB / N. |

This is a clear generalization of the standard procedure of dividing a segment into two (N = 2).

### References

- E. Maor,
*The Pythagorean Theorem*, Princeton University Press, 2007

- How to divide a segment into n equal parts
- Al-Nayrizi's Construction
- Besteman's Construction
- Besteman Construction II
- Dividing a Segment by Paper Folding
- Euclid's Segment Division
- The GLaD Construction
- The SaRD Construction
- Similar Right Triangles
- Divide Triangle by Lines Parallel to Base

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Copyright © 1996-2018 Alexander Bogomolny63592921 |